now y = cos(theta)
=> y = cos(arcsin(x/4))
now recall that arcisn(x/4) = theta and sin(theta) = x/4
so we draw a right triangle with one acute angle being theta, the length of the side opposite to the angle is x and the hypotenuse is length 4. by pythagoras, the adjacent side is sqrt(16 - x^2), from this triangle we see that:

well what i did was x=tan(t) y=sec(t)
1+tan^2=sec^2
1+x^2=y^2
y^2-x^2=1
hyperbola
similarily x=4sin(@)
x/4=sin(@)
sin^2+cos^2=1
(x/4)^2+y^2=1 same answer though can you help me with
number 8 and second part of number 9

well what i did was x=tan(t) y=sec(t)
1+tan^2=sec^2
1+x^2=y^2
y^2-x^2=1
hyperbola
similarily x=4sin(@)
x/4=sin(@)
sin^2+cos^2=1
(x/4)^2+y^2=1 same answer though can you help me with
number 8 and second part of number 9

that's nice! i never thought of using identities. i should remember that technique for the future

question number 8 doesnt really make sense isnt it suppose to be polar coordinates? what i tried to do for xy=1 is uhmm x and y has to rather equal pi/4 or 5pi/4(cuz its int he 3rd quardnant) from there i did tan(@)=1 @=pi/4 or 5pi/4 x=rcos x=1 therefore 1=rcos(@) 1=r(1/sqrt2) r=+-sqrt2 and for sin i got pi/4+npi

and for 9 how did you get one half can you walk me through that problem again, it sort of doesnt make sense

all i did was complete the square for y, do you see it now? i had to add the square of half the coefficient of y, and because we're dealing with an equation, i had to add it to the other side as well, to keep things equal